Draw the reactants and products of the following reaction:
3-ethyl-2-heptene + hoh → 3-ethyl-3-heptanol + 3-ethyl-2-heptanol
consider how it is possible for two products to form in this reaction, and identify the major product, giving your reasons for both.
what type of reaction is this classified as?

This reaction is an hydration of alkenes and the major product is 3-ethyl-2-heptanol

Explanation:

In this reaction in the first step water is the nuclophile and the alkene is the electrophile. Then, the carbanion is the nucleophile and attacks a proton from the water forming the respective alcohol. This carbanionic compound is a reaction intermediary. You can see the mechanism in the picture.

Is that carbanionic compound who explained the major product because the more substituted carbanion the most stable intermediary. Thus, the 3-ethyl-2-heptanol intermediary is the most stable and this product is the major one.

I hope it helps!

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given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g