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Question

In the following sequence, product I, J, and L are formed. K represents a reagent.


The structure of the product L is:
 



Solution

The correct option is D


Let us look at the reagents preceding I. We have an aldehyde, which will get reduced to the corresponding alcohol (the triple bond is unaffected). Further, the alcohol reacts with PBr3 to yield the corresponding alkynyl halide.

Getting to I is pretty straightforward. The alkynyl halide – when treated with Mg in dry ether, forms the Grignard reagent. The alkynyl part of the Grignard reagent may be (an incorrect but useful approximatin) thought of a carban ion – which on carboxylation plus workup gives the corresponding carboxylic acid with one more carbon than before.

In the next step, thionyl chloride (phosphorous chlorides could also be used) yields the corresponding acid chloride.

Look at the catalyst in the final step: it is Palladium supported on Barium Sulphate but poisoned by quinoline. Most of you might instantly recognize that it is the Rosendmund catalyst. It is also known as the Cram-Allinger catalyst or Lindlar Catalyst. It reduces the acid chloride to aldehyde and the triple bond to a cis-alkene. Hence the product:

 

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