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1 answer:
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Answer:
21.2 moles.
Explanation:
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In this case, for the given chemical reaction, we can see there is a 1:4 mole ratio between tetraphosphorous decaoxide and phosphorous; therefore, the following proportional factor provides the requested moles of phodphorous:
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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 () 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.
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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.
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