the general formula for alkyl radicals is *
1 point
CnH2n
CnH2n+2
CnH2n_1
CnH2n+1
Answers
Answer:
B option
Explanation:
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Step 1. To determine the molecular weights of compounds (B) and (C).
(i) The mol. wt. of compound (A) is 54 while that of compound (B) which it gives on treatment with an excess of Br
2
in CCl
4
is 593% more than that of A.
∴ Mol. wt. of (B)=
100
(100+593)
×54=374.22
Thus, the increase in weight due to addition of Br atoms =374.22−54.0=320.22
Since the atomic weight of Br is 80, therefore, the number of Br atoms added =320.22/80=4
Therefore, the hydrocarbon (A) must be an alkyne.
(ii) Further since the mol. wt. of compound (C) which hydrocarbon (A) gives on catalytic hydrogenation is only 7.4% more than that of (A), therefore, the mol. wt. of (C)
=
100
(100+7.4)×54
=57.994=58(approx.)
Thus, the increase in weight due to the addition of H-atoms =58−54=4
Since the atomic weight of H is 1, therefore, the number of H - atoms added during catalytic hydrogenation =4/1=4
Therefore, the hydrocarbon (A) must be an alkyne.
Step 2. To determine the structure of the hydrocarbons (A), (B), (C) and (D).
(i) The two possible structures for the hydrocarbon (A), i.e. an alkyne with mol. wt. 54 (M.F. C
4
H
6
) are:
But-1-yne (I)
CH
3
C
2
−C≡CH
But-2-yne (II)
CH
3
−C≡C−CH
3
(ii) Since (A) to reacts with CH
3
CH
2
Br in presence of NaNH
2
to give another hydrocarbon (D), therefore, (A) must be a terminal alkyne, i.e. but-1-yne and not but-2-yne.
(iii) If hydrocarbon (A) is but-1-yne, then the structure of the compounds (B), (C) and (D) may be worked out as follows: (refer to image)
Step 3. To determine the structure of the diketone (E).
Since the hydrocarbon (D) i.e. hex-3-yne on ozonolysis gives a diketone (E) which on further oxidation gives propionic acid, therefore, the diketone (E) must be hexane-3, 4-dione as explained below:
Thus, (A) = but-1-yne, (B) = 1,1, 2, 2-tetrabromobutane, (C) = butane, (D) = hex-3-yne and (E) = hexane -3, 4-dione.