Compound A and compound B are constitutional isomers with molecular formula C3H7Cl. When compound A is treated with sodium metho
1 answer:
Answer:
Compound A and compound B are constitutional isomers with molecular formula C3H7Cl.
When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination reaction predominates.
Explanation:
Constitutional isomers are the one which differs in the structural formula.
When compound A is treated with sodium methoxide, a substitution reaction predominates.
That means sodium methoxide is a strong base and a strong nucleophile.
But when it reacts with primary alkyl halides it forms a substitution product and when it reacts with secondary alkyl halide it forms mostly elimination product.
The reaction and the structures of A and B are shown below:
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Answer: The bond between boron and hydrogen in boron trihydride is covalent bond.
Explanation:
The type of bonding between the atoms forming a compound is determined by using the electronegativity difference between the atoms. According to the pauling's electronegativity rule:
- If
, then the bond is non-polar.
- If
, then the bond will be covalent.
- If
, then the bond will be ionic.
We are given:
Electronegativity for boron = 2.0
Electronegativity for hydrogen = 2.1
As, is less than 1.7 and not equal to 0. Hence, the bond between boron and hydrogen is covalent bond.