Match the chemical substances in Column - I with type of polymer- type of bonds in Column - IIP - text Starch -1 - text Natural Polymer hlineS - text Nylon- 10-6-4 - text Glycoside linkage-5 - text Disaccharides line endarray Match each product in LIST-I with one or more reaction in LIST-II and choose the correct option- A- P - 3-4 - Q - 1-2 - R - 1-2-4 - S - 3-5B- P - 1-4 - Q - 1-3 - R - 4-5 - S - 2-3C- P - 3-5 - Q - 2-4 - R - 1 - S - 4D- P - 2-3-4-5 - Q - 2-3-4 -5 - R - 1-2-4 - S - 1-2

The correct option is C P → 3.5; Q → 2,4; R → 1; S → 434 No of H= 16, no of π electrons= 14, no of lone pair electrons= 4 so 16+14+4 =34 ...

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Byju's Answer

Standard XII

Chemistry

Hyperconjugation

Match the che...

Question

Match the chemical substances in Column – I with type of polymer/ type of bonds in Column - II
$\begin{matrix} Columns - I & Column - II (P) & Starch & (1) & Natural Polymer (Q) & Protein & (2) & Synthetic polymer (R) & Maltose & (3) & Amide linkage (S) & Nylon-10,6 & (4) & Glycoside linkage (5) & Disaccharides \end{matrix}$

Match each product in LIST-I with one or more reaction in LIST-II and choose the correct option.

P \to 3.4; Q \to 1, 2; R \to 1, 2, 4; S \to 3, 5

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P \to 1.4; Q \to 1, 3; R \to 4, 5; S \to 2, 3

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P \to 2, 3.4, 5; Q \to 2, 3.4.5; R \to 1, 2, 4; S \to 1, 2

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P \to 3.5; Q \to 2, 4; R \to 1; S \to 4

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Solution

The correct option is D $P \to 3.5; Q \to 2, 4; R \to 1; S \to 4$
34

No of H= 16, no of π electrons= 14, no of lone pair electrons= 4 so 16+14+4 =34

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The correct option is D P→3.5;Q→2,4;R→1;S→434 No of H= 16, no of π electrons= 14, no of lone pair electrons= 4 so 16+14+4 =34

The correct option is D $P \to 3.5; Q \to 2, 4; R \to 1; S \to 4$
34

No of H= 16, no of π electrons= 14, no of lone pair electrons= 4 so 16+14+4 =34