Benzyl bromide, on treatment with a nucleophile - triphenylphosphine, is then made to react with b-Butyl lithium - a strong base to give a yhosphorus ylid.
Which of the following would yield the hydrocarbon:

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phenylacetaldehyde

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None of the above

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Solution

The correct option is B

The first part - the question statement generates an ylid (phosphorus).
This can olefinate (make an alkene) the carbonyl carbon, replacing the double bond with O
with a double bond with carbon.

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Benzyl bromide, on treatment with a nucleophile - triphenylphosphine, is then made to react with n-Butyl lithium - a strong base to give a phosphorus ylid.
Which of the following, on treatment with the said ylid - would yield the hydrocarbon:

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(R) Sodium methoxide is a strong nucleophile.

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Benzyl bromide, on treatment with a nucleophile - triphenylphosphine, is then made to react with b-Butyl lithium - a strong base to give a yhosphorus ylid. Which of the following would yield the hydrocarbon:

The correct option is B The first part - the question statement generates an ylid (phosphorus). This can olefinate (make an alkene) the carbonyl carbon, replacing the double bond with O with a double bond with carbon.

Benzyl bromide, on treatment with a nucleophile - triphenylphosphine, is then made to react with b-Butyl lithium - a strong base to give a yhosphorus ylid.
Which of the following would yield the hydrocarbon:

The correct option is B

The first part - the question statement generates an ylid (phosphorus).
This can olefinate (make an alkene) the carbonyl carbon, replacing the double bond with O
with a double bond with carbon.