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3 years ago

The molecules CH4 and CH3Cl both have tetrahedral shapes. Why is CH4 nonpolar whereas CH3Cl is polar?

2 answers:

6 0

Because in CH4 the valence electrons of Carbon are being pulled away from the nucleus with the same force (since all hydrogen has the same electronegativity).So charges are equally distributed throughout the atom and the atom is non-polar. In CH3Cl Cl has a different electronegativity than H, so charges aren't equally distributed and therefore the molecule is polar.

vova2212 [387]3 years ago

3 0

Answer is:

1) in molecule of methane (CH₄) all the atoms around the central element (carbon) are equivalent (four hydrogens) and it has a net dipole moment of 0 (vectors of dipole moments cancel each other, dipole moment is zero).

2) in molecule of chloromethane (CH₃Cl) there are different elements (one chlorine and three hydrogen atoms) surrounding the central atom (carbon) and it has higher net dipole moment.

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Give the structure of the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane.

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the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

what is free radical halogenation?

A substitution reaction in which a hydrogen atom is replaced with a halogen atom, via a free radical reaction mechanism. when this reaction is carrid out by bromine radical, it is called free radicle bromination. When bromine ( $Br{2}$ ) treated with light (hν) it comes to homolytic cleavage of the Br-Br bond and give rise to bromine radicles.

free-radical bromination of 2,2,4−trimethylpentane

Bromination of an alkane includes the substitution of a bromine atom for a hydrogen atom. The following stages will be taken by 2,2,4-trimethylpentane during this reaction:

Initiation reaction: The production of a bromine free radical requires the initiation of heat or light.

$Br - Br$ ⇒ $2Br$ ·

Propagation: This reaction relies heavily on hydrogen. This reaction is impossible if hydrogen is not present. Because tertiary free radicals are more stable than secondary and primary free radicals, they are favoured in this reaction.

Termination: The remaining free radical of bromide reacts with the tertiary free radical of 2,2,4-trimethylpentane to form 2-bromo-2,4,4-trimethylpentane.

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

To know more about free radical halogenation, check out:

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