CCl 3- C - C - H D 2 - Pd - HaSO 4- A HCl - BCompound A and B respectively are-A-B-C-D-

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Standard XII

Chemistry

Solubility Product

CCl 3- C ≡ C ...

Question

$C C l_{3} - C \equiv C - H D_{2} / P d - H a S O_{4} - -------- \to A H C l - - \to B$
Compound A and B respectively are:

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Solution

The correct option is B

Secondary carbocation will be highly destabilized by electron withdrawing power of $C C l_{3} -$ hence here it forms primary carbocation and result in antimorkonikov product.

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CCl3−C≡C−HD2/Pd−HaSO4−−−−−−−−−→AHCl−−→B Compound A and B respectively are:

The correct option is B Secondary carbocation will be highly destabilized by electron withdrawing power of CCl3− hence here it forms primary carbocation and result in antimorkonikov product.

$C C l_{3} - C \equiv C - H D_{2} / P d - H a S O_{4} - -------- \to A H C l - - \to B$
Compound A and B respectively are:

The correct option is B

Secondary carbocation will be highly destabilized by electron withdrawing power of $C C l_{3} -$ hence here it forms primary carbocation and result in antimorkonikov product.