Question

Question 43 (2 points) A triple bond is composed of Oa) Two sigma and one pi bond ob) Three sigma bonds Oc) Three pi bonds od) One sigma and two pi bonds Question 44 (2 points) Identify the major product for the following reaction. (CH) CHCH.C=CH HO, H2SO4 HgSO4 OH III IV Oa) IV ob) Oc) II od) III

Answer 1

Qn 43:

We know that the triple bond has one sigma bond and two pi-bonds.

The electronic configuration of Carbon is 1S2 2S2 2P2 There are four electrons in its outer most shell.

On exitation it forms 2S1 2P3 configuration.

In triple bonds, the Carbon is SP hybridised utilising one each electron from S and P orbital, leaving two unpaired P- orbital electrons.

The SP hybridised orbital forms the sigma bond. The two unpaired electrons of P orbital forms the two pI bonds.

Thus the option is d) one sigma and two pi bonds.

Qn 44:

The alkyne under goes addition of H2O on the unsaturation forming a ketone.

The formation is explained as below:

H2O CH3 CHB -CH-CH2-c= CH2-c = CH2 CH-CH₂-C=CH H2SO4 Hg sou CH3 CH3 H2o Hasol) Egsoy OW CHB -J DA CH lp •CH - CH2- CH3 CA-CH₂-C es - 420 CH3 CH3 CH 11 [1]

The addition of two molecules of H2O on the alkyne gives rise to an unstable gem diol which releases water to form the product which is identical to the structure given by compound [ I ].

Hence the option is b) I

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