Unlike E2 reactions, in which hydrogen abstraction occurs simultaneously with the dissociation of the leaving group (limiting the configuration of the reaction's product), E1 reactions occur in two distinct steps. The slow rate-determining step that must first occur is the dissociation of the leaving group. Leaving behind a carbocation intermediate, it is often necessary to consider possible carbocation rearrangements that would stabilize the positive charge.

In this case, no such rearrangement is favourable as there are no locations of greater stability available.

However, what must be considered is that the intermediate is free to orient itself in its most stable conformation prior to the formation of the double bond in the second step. As a result, the E product (the larger substituents are oriented opposite one another with respect to the double bond) is yielded primarily.

In order to convert an alkene into an alkane, we need a Pd catalyst

Potassium permanganate is a strong oxidizing agent. It can convert secondary alcohols to ketones. It can also convert primary alcohols to carboxylic acids. 1-propanol has a hydroxy group on carbon 1, so it is primary; thus it will be converted to propanoic acid.

The reagents are indicative of the formation of an epoxide by mCPBA (m-chloroperoxybenzoic acid), followed by epoxide-opening using methylmagnesium bromide (Grignard reagent, a nucleophile). Epoxidation can occur from either side of the alkene (in this instance, epoxidation is shown from the top face). On treatment with the methyl Grignard reagent, backside attack occurs (in this instance, the nucleophile approaches from the back of the molecule), with nucleophilic addition to the least hindered epoxide carbon (steric effects).

Aromatic compounds follow four rules: (1) They are conjugated—there needs to one “p” orbital from each atom in the ring, so each atom must be either sp² or sp hybridized; 2) They are cyclic: linear systems are not aromatic; 3) They are planar: there is good overlap/interaction between p orbitals; 4) they follow the The Huckel Rule—4n+2 π electrons in the cyclic conjugated π system (n = 0, 1, 2, 3 etc.). Of the possibilities,compounds I (14 e⁻, n = 3) and III (10 e⁻, n = 2) obey the rule.

This is a little advanced and at the top of the difficulty the DAT can throw at you, you have to consider compound 3 as one molecule, not two separate rings. Aromaticity requires a planar conjugated system, so we can ignore the bridge carbon sticking out of the plane of the rest of the molecule. There cannot be any sp3 atoms IN the aromatic ring. The carbon bridge actually pushes the 10 carbon ring to be planar, without it the molecule wouldn’t be aromatic.

The answer is "Internal alkynes are more stable than terminal alkynes" as it is the only true statement in regards to alkynes. Internal alkynes are more stable because they have a better conjugated system than terminal alkynes. A conjugated system is a system of a single bond, then a multiple bond, then a single bond, and so on. A conjugated system will always be more stable than an unconjugated system. It is evident that the internal alkyne follows the conjugated system and the terminal alkyne does not based on the picture below.

Two sets of reagents are required to convert butanone into 2-butene. First, we use NaBH4 to reduce the butanone into a 2-butanol. Second, we use heat and acid to dehydrate the butanol and yield the final desired product.

1. H3CMgBr; 2. Heat/H3O+ may seem like an acceptable answer choice. However, note that the Grignard reagent converts the butanone into a tertiary alcohol, rather than a secondary alcohol as needed.

When comparing relative volatilities of compounds, you must consider the molecular weight of a compound, as well as the intermolecular attractive forces between the identical molecules found in a sample of the compound in question.

We can eliminate choices I, II and V. These compounds have functional groups that feature polarized X-H bonds, allowing molecules in a sample of these compounds to participate in hydrogen bonding. Hydrogen bonding is a strong attractive force, and thus more energy would have to be put into a sample to vaporize it (boil a liquid sample). In other words, the hydrogen bonds will raise the boiling point and lower the volatility of these compounds.

Answer choice IV, which features an alkyl bromide, may also be eliminated for two reasons. First, as bromine is much heavier than carbon, molecule IV will be much heavier than III, and will thus require much more energy to transition into the gaseous state. Secondly, as bromine is fairly electronegative, the molecule will feature a dipole in the carbon-bromine bond, and thus a sample of IV will experience dipole-dipole attractive interactions. As described above, attractive intermolecular interactions require more energy to overcome in order for a sample to undergo a liquid-gas phase change. Thus, molecule IV is less volatile than molecule III, the correct answer.

S_N2 reactions involve a backside nucleophilic attack on an electrophilic carbon. As a result, less steric congestion for this backside attack results in a faster reaction, meaning that S_N2 reactions proceed fastest for primary carbons. In addition, beta-branching next to a primary carbon results in a slower reaction, as does a poorer leaving group (i.e. chloride instead of bromide).

1-bromopentane has a good leaving group (bromine) attached to a primary carbon with no beta-branching, meaning it will proceed the fastest.

PCC is an oxidizing agent. It converts alcohols to ketones, but is not strong enough to convert primary alcohols to carboxylic acids. 2-butanol has a hydroxy group on its carbon 2. The addition of PCC will convert this hydroxy group into a carbonyl, producing 2-butanone.

The longest carbon chain that can be formed is eight carbons. The base molecule is an octane.

Using IUPAC rules, substituents should have the lowest possible numbers; thus, we start counting carbons from the right side rather than the left. If you count from the correct side, there are two methyl groups on carbon 3 and one on carbon 5. Thus, the name of the molecule is 3,3,5-trimethyloctane.

There are 14 pi electrons because oxygen must contribute 2 pi electrons to avoid antiaromaticity.  The other 12 pi electrons come from the 6 double bonds.

Nitrogen does not contribute any pi electrons, as it is sp2 hybridized and it's lone pairs are stored in sp² orbitals, incapable of pi delocalization.  Each nitrogen's p orbital is occupied by the double bond.

Hydrogenation of an alkene with a metal catalyst, such as platinum, occurs via syn addition.

It is important to note the three main types of reactions for alkenes. The first type of reaction is a 2-step mechanism in which the electrophile attacks the carbocation nucleophile. This can yield syn or anti-products. The second type of reaction is a 2-step mechanism that forms a bridged carbocation as the intermediate. This can yield only anti products. The third and last type of reaction is a 1-step addition. This can only yield syn products.

An example of the third type of reaction is the addition of hydrogen with palladium, platinum, or nickel as demonstrated in the picture.

The rate of an SN1 reaction is determined only by the concentration of the substrate. Unlike an SN2 reaction, where the addition occurs in one step and requires the activity of the substrate and the nucleophile, an SN1 reaction occurs in two steps and is only limited by the activity (i.e. leaving ability) of the substrate. Once the leaving group leaves the substrate, the nucleophile does not hesitate to attack the exposed carbocation.

A nucleophile acts by donating a pair of electrons to another atom's nucleus. In general, a negatively charged compound is going to be a stronger nucleophile than a neutral compound. In addition, as one proceeds down a given column of the periodic table, the nucleophilicity increases because the electrons are not held as tightly to the nucleus (electronegativity decreases).

CH3S−is the best nucleophile, because it has a negative charge (more electron density), and its electrons are held less tightly than those of CH3O−because sulphur is less electronegative than oxygen.

When converting a linear sugar to its ring form, a bond is formed between the oxygen attached to carbon 5 and the carbon at position 1. All hydroxyl groups that are not attached to the carbon in position 1 and are oriented to the right end up trans to the CH2OH attached to carbon 5, while those that are in the left position end up cis to the CH2OH attached to carbon 5.

If the hydroxyl group attached to carbon 1 ends up trans to the CH2OH attached to carbon 5, the ring structure is considered alpha. If the hydroxyl group attached to carbon 1 is cis to the CH2OH attached to carbon 5, the ring structure is considered beta.

The alpha ring structure of D-mannose bonds the carbon 1 hydroxyl group trans to the carbon 5 CH2OH group. The hydroxyl groups on carbons 2, 3, and 4 will be cis, cis, and trans with respect to the CH2OH.

The S_N1 mechanism involves the formation of a carbocation intermediate in the rate-determining step. The most stable carbocation will produce the fastest reaction. We can immediately eliminate any answer choices that will produce primary or secondary carbocations since a tertiary carbocation will be much more stable. When comparing tertiary carbocations, larger and more electronegative substituents will allow for more charge stabilization.

Since the tertiary carbocation formed by the dissociation of iodide from (CH3)3CI will the be most stable, this substrate will react the fastest.

A: S; CH(C)2 > CH2CH(C)2 > CH2CH2 > H, here H is pointing out, so we should look from the back side.
B: R; Si > O > C > H, also we should look from the back side.
C: R; O > C=O > C(C)3 > H
D: R; O > C(C)3 > CH3 > electron pair
E: This isn’t chiral, it has a plane of symmetry!

In this question, we're told that our starting material, 1-hexene, is being reacted with hydrobromic acid in the presence of ultraviolet (UV) light. To solve this, we need to consider how halogens add to alkenes, specifically in the presence of ultraviolet light.

First, it's important to note that UV light will cause the hydrogen and bromine atoms in HBr to dissociate as free radicals. Because each of these atoms is electron deficient, they desperately want to react in order to fill their valence shells. And since the double bond in the alkene is electron-dense, a reaction will occur between one of the electrons in this double bond and the free radical bromine.

The bromine radical adds to the alkene first at the 1-carbon in an anti-Markovnikov fashion. The reason it adds to this carbon (and not the 2-carbon) is that a secondary free radical is more stable than a primary free radical. Upon the formation of this secondary radical, it will react with the hydrogen-free radical in solution to generate the finished product, 1-bromohexane.

The above reaction would more readily proceed if the electrophilicity of the carbonyl carbon were enhanced. This may be achieved through electron withdrawal via the R group.

The ether (-OMe), the methyl (-Me), and the hydroxyl (-OH) would all produce an electron-donating effect and are thus incorrect answers.

The nitro group (-NO₂), and the positively charged, tetra-substituted amino group (consider the structure once this trimethyl amino group is connected to the aryl ring) are both electron-withdrawing. As the trimethyl amino group will have an overall positive charge (and the nitro group is neutral overall), the trimethyl amino group is the stronger electron-withdrawing moiety and is thus the correct answer.

Whenever an alkane is reacted with excess oxygen in a high-temperature reaction, the reaction is known as combustion. Combustion of an alkane always produces carbon dioxide gas and water vapour.

Therefore the correct answer is CO2 and H2O.

A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor.

The carbocation is clearly trying to accept electrons due to the positive charge. ZnCl2 is ionic and neutral. The boron in BF3 has only three electrons, and they are all creating bonds with fluorines.

The correct answer has a lone pair on the nitrogen, and thus has electrons to donate and as a Lewis base. H3CCN is not a Lewis acid.