1.

0.0852g of an organic halide (A) when dissolved in 2.0g of camphor, the melting point in the mixture was found to be 167^(@)C. Compound (A) when heated with sodium gives a gas (B). 280mL of gas (B) at STP weight 0.375g. What would be 'A' in the whole process? K_(f) for comphor=40, m.pt. of camphor=179^(@)C.

Answer»

`C_2H_5Br`
`CH_3I`
`(CH_3)_2CHI`
`C_3H_7Br`

Solution :`DeltaT=179-167=12, w=0.0852g, W=2g, K_f`=40, molecular WEIGHT of (A)
`=(1000xxK_fxxw)/(DeltaTxxW)=(1000xx40xx0.0852)/(12xx2)=142`
(A) undergoes Wurtz REACTION to form (B) i.e.
`(A) overset(Na)to(B)+NaX`
(B)is an ALKANE say `C_nH_(2n+2)`
`:'` 280 mL of (B) weighs 0.375 g at NTP
`:.` 22400 mL of (B) weighs `=(0.375xx22400)/280`
= 30 g at NTP
`:.` M.wt. of (B) =30,12n+2n+2n+2=30,n=2
Thus (B) si ethane and THEREFORE (A) is `CH_3X`.
The m.wt.of `CH_3X=142`
At. wt. of X = 127 `:.` X is iodine
Therefore alkyl ahlide is `CH_3I`. This reaction is `underset((A))(2CH_3I)underset("either")overset(Na)tounderset((B))(C_2H_6)`


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