Question 5 Not yet answered Marked out of 5.00 Consider the following sequences

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Question 5 Not yet answered Marked out of 5.00 Consider the following sequences of the reaction. Match the product name on each step with the letter that represents this product 1. BHs 2. NaOH, H,O Flag question Br. heat or light Br HBr NaOEt Br. reflux A Choose... Choose B (2S,3S)-2,3-dibromobutane 1-pentene C 1-butene x(2R,3R)-2,3-dibromobutane Y 2-bromopentane (2R,3S)-2,3-dibromobutane 1-pentanol 2-bromobutane (2S,3S)-2,3-dibromobutane and (2R,3R)-2,3-dibromobutane (2R)-1,2-dibromobutane and (2S)-1,2-dibromobutane trans)-but-2-ene G Choose... V Choose M Choose...

Explanation / Answer

Q 5.

A

B

C

A :SN1 MECHANISM FOR REACTION OF ALCOHOLS WITH HBr

Step 1:
An acid reaction. Protonation of the alcoholic oxygen to make a better leaving group. This step is very fast and reversible. The lone pairs on the oxygen make it a Lewis base.  

Step 2:
Cleavage of the C-O bond allows the loss of the good leaving group, a neutral water molecule, to give a carbocation intermediate. This is the rate determining step (bond breaking is endothermic)  

Step 3:
Attack of the nucleophilic bromide ion on the electrophilic carbocation creates the alkyl bromide.  

B: when t-butoxide is used, it will preferentially remove the proton from the smaller group. This produces the so-called “Hoffmann” product

C: Hydroboration-oxidation transforms alkenes into alcohols. It performs the net addition of water across an alkene. The reaction begins with the concerted syn addition of B and H across the double bond, with the boron adding to the less substituted carbon In the second step, hydrogen peroxide and a base such as NaOH are added. the NaOH deprotonates the hydrogen peroxide which makes the conjugate base of hydrogen peroxide (a better nucleophile than H2O2 itself). The resulting NaOH then attacks the boron This sets up the key migration step, where the carbon-boron bond migrates to the oxygen bound to boron, breaking the weak oxygen-oxygen bond The OH expelled then comes back to form a bond on the boron resulting in the deprotonated alcohol (alkoxide). The alkoxide is then protonated by water or some other comparably acidic species

(Note: The World File is pest as attachment.)

X: In the presence of light, or at high temperatures, alkanes react with halogens to form alkyl halides. Reaction with bromine gives an alkyl bromide.

Y: when t-butoxide is used, it will preferentially remove the proton from the smaller group. This produces the so-called “Hoffmann” product

Z: Unsaturated hydrocarbons such as alkenes and alkynes are much more reactive than the parent alkanes. They react rapidly with bromine, for example, to add a Br2 molecule across the C=C double bond.

X

Y

Z

G:  two products can form when an unsymmetric reagent such as HBr is added to an unsymmetric C=C double bond. In practice, Markovnikov's rule states that the hydrogen atom adds to the carbon atom that already has the larger number of hydrogen atoms when HX adds to an alkene.

V:  Treatment with the strong base sodium ethoxide (NaOEt) gives two alkenes The trans product dominates over the cis product (due to less steric crowding) and more substituted product formed.

Q 6.

Initiation Step:

The reaction begins with an initiation step, which is the separation of the halogen (X2) into two radicals (atoms with a single unpaired electron) by the addition of uv light. This is called the initiation step because it initiates the reaction.

Propogation Steps:

The initiation step, or the formation of the chlorine radicals, is immediately followed by the propogation steps--steps directly involved in the formation of the product. As an example, isobutane (C4H10) will be used in the chlorination reaction. The first step is the abstraction of the hydrogen atom from the tertiary carbon (a tertiary carbon is a carbon that is attached to three other carbon atoms) Note that these are not protons (H+ ions) that are being abstracted, but actual hydrogen atoms since each hydrogen has one electron. This first propogation step forms the tertiary radical.

In the last step, the tertiary radical then reacts with another one of the Bromine molecules to form the product. Notice that another bromine radical is regenerated, so this reaction can, in theory, go on forever as long as there are reagents. This is called a chain reaction.

Termination Steps:

Side reactions that can stop the chain reaction are called termination steps. These termination steps involve the destruction of the free-radical intermediates, typically by two of them coming together.

A

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