Reactants

Question 1

KOC(CH3), NaOEt

Answer 1

:OC(CH3), :OEt Will steal Hydrogens from hydrocarbon chains to form double bonds (alkenes)
Also can form bonds to Carbons if there is an available leaving group (Br)

Question 2

B2H2

Answer 2

Anti Markovnikov -OH (alcohol addition)

Question 3

NaI

Answer 3

I- will replace Br- and Na+ will bind with Br

Question 4

Na2Cr2O7

Answer 4

Cr 6+ will always form acetic acid O=C-OH with alkyl carbon

Question 5

How to make -OH a leaving group?

Answer 5

Add HBr. The -OH will grab an H, become charged and leave, allowing Br to take its place on the hydrocarbon chain.

Question 6

Ozone

Answer 6

Splits double and triple bonds, turning loose ends into acetic acids.

Question 7

NaNH2

Answer 7

Powerful base that will take Hydrogens, even from triple bonded Carbons (alkynes)

Question 8

SOCl2 and pyridine

Answer 8

With primary or secondary alcohols will produce R-Cl, SO2 and HCl

Question 9

Dibromination Alkane

Answer 9

Free radical. Will add 2 Br’s to the tertiary Carbons.

Question 10

Hydroboration (B2H6)

Answer 10

Breaks double bond of alkene to form anti markovnikov alcohol. Because Boron is more electropositive than H and therefore influences the H position

Question 11

Dibromination Alkene

Answer 11

Forms triangular bromonium ion intermediate over the double bond and double bond breaks with trans addition of the second Bromide.

Question 12

Adding Hg Alkene

Answer 12

A Markovnikov alcohol addition

Question 13

Double bond O C-O-O-H Alkene

Answer 13

Epoxidation, secondary O will break double bond and form an epoxide over it.