GRE Subject Test: Chemistry : Intermediates

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

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Example Questions

Example Question #201 : Gre Subject Test: Chemistry

What intermediate is involved in the conversion of compound B to compound C?

Mcat_1

Possible Answers:

Tertiary carbanion

Tertiary carbocation

Tertiary radical

Secondary carbocation

Secondary radical

Correct answer:

Tertiary carbocation

Explanation:

The strong sulfuric acid protonates the hydroxyl group of compound B, resulting in the loss of water as a leaving group and the generation of a carbocation intermediate. Since this carbocation carbon is attached to three other carbons, this is a tertiary carbocation. It is bound to the phenyl substituent, a methyl group, and the branched carbon chain.

Example Question #211 : Gre Subject Test: Chemistry

Which of the following carbocation intermediates requires the least activation energy?

Possible Answers:

Cannot be determined

Carbo3

Carbo4

Carbo2

Carbo1

Correct answer:

Carbo1

Explanation:

The more stable the carbocation, the lower the activation energy for reaching that intermediate will be. The more substituted a carbocation is, the more stable it is. The carbocation bonded to three alkanes (tertiary carbocation) is the most stable, and thus the correct answer.

Secondary carbocations will require more energy than tertiary, and primary carbocations will require the most energy.

Example Question #211 : Gre Subject Test: Chemistry

Carbon 1:

8

Carbon 2:

9

Let's say we react the given compound with . During the first step of the reaction, will the hydrogen be added to carbon 1 or carbon 2, why?

Possible Answers:

Carbon 1 because the intermediate will then be a secondary carbocation

Carbon 2 because the bromine will first attack the double bond from the least sterically hindered side

Carbon 1 because the intermediate will be a primary carbocation

Carbon 1 because the bromine anion is large and requires space to react

Carbon 2 because the intermediate will be a secondary carbocation

Correct answer:

Carbon 1 because the intermediate will then be a secondary carbocation

Explanation:

The correct answer is: carbon 1 because the intermediate will then be a secondary carbocation.

This is a case of  addition across a double bond where  stands for any halide. The first step in this reaction is the attack of  by the double bond. This will create two intermediates, the first being the halide anion  (so in our case ), the second being a carbocation on our compound at one of the two carbons that formerly shared the double bond. 

If the hydrogen attached the carbon 2 then we would have a positive charge on carbon 1 and vice versa. A positive charge on carbon 1 is known as a primary carbocation (a carbon attached to 1 other carbon or function group), which is rarely if ever seen due to its overwhelming instability. A positive charge on carbon 2 is known as a secondary carbocation (a carbon attached to 2 other carbons or functional groups) and is much more stable than a primary carbocation.

We would want the more thermodynamically stable intermediate for our reaction to proceed, so we would want the positive charge on carbon 2 and the hydrogen attached to carbon 1. 

Example Question #1 : Intermediates

4

 

If the carbon being pointed to was deprotonated (resulting in a positive charge on it). Would the resonance form (the positive charge being redistributed to the carbon with a bromine) be more stable than a secondary carbocation? Why? 

Possible Answers:

No because bromine's bulky electron cloud will interfere

Yes because bromine donates some electrons

Yes because bromine is an electrophile

No because bromine is an electrophile

No because secondary carbocations are unstable

Correct answer:

No because bromine is an electrophile

Explanation:

The resonance form of this compound would put the positive charge on the carbon attached to the bromine. Unfortunately this carbon is already slightly positive due to the electron withdrawing effects of bromine due to its high electrophilicity. So this resonance form would be more unstable than a secondary carbocation due to the increased concentration of positive charge from bromine's electron withdrawal. 

Example Question #33 : Organic Chemistry, Biochemistry, And Metabolism

Which of the following steps of free radical chlorination does not produce a free radical as a product?

Possible Answers:

Termination

Propagation

Initiation

Halogenation

Correct answer:

Termination

Explanation:

The three steps of a free radical chlorination reaction are, in order, initiation, propagation, and termination.

Free radicals are produced in the initiation and propagation steps. The termination steps combine any two free radicals formed in the reaction to produce a compound that has no unpaired electrons (free radicals).

Example Question #1 : Intermediates

Which of the following transformations includes an enolate intermediate?

 Q12

Possible Answers:

II and III

I, II, and III

III only

I and II

I and III

Correct answer:

II and III

Explanation:

Enolates are formed by an oxygen anion bound to an alkene carbon. Reactions II and III include an enolate intermediate, as shown in the mechanisms below, whereas reaction I is a simple SN2 reaction and does not include an enolate intermediate. Enolates are highlighted in red. 

A12

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