Pent-4-enenitrile, H2C=CH-CH2-CH2-CN\text{H}_2\text{C=CH-CH}_2\text{-CH}_2\text{-CN}H2C=CH-CH2-CH2-CN, can be used as a starting material for the synthesis of hexane-1,6-diamine, as shown in this reaction scheme:
H2C=CH-CH2-CH2-CN→HBrReaction 1Br-CH2-CH2-CH2-CH2-CN(Isomer Y) \text{H}_2\text{C=CH-CH}_2\text{-CH}_2\text{-CN} \xrightarrow[\text{HBr}]{\text{Reaction 1}} \text{Br-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CN} \quad (\text{Isomer } \mathbf{Y}) H2C=CH-CH2-CH2-CNReaction 1HBrBr-CH2-CH2-CH2-CH2-CN(Isomer Y)Use IUPAC rules to name isomer Y\mathbf{Y}Y.
Reaction 1 produces a mixture of Y\mathbf{Y}Y and two other isomers. Draw the structures of these two other isomers and explain, by considering the mechanism of this reaction, why all three isomers are formed.
Identify the reagent and condition needed for Reaction 2.
Identify the reagent and reaction conditions needed for Reaction 3, and write a balanced equation for it.
An incomplete equation for the formation of nylon 6,12 from eight molecules of hexane-1,6-diamine and eight molecules of dodecanedioic acid is shown below:
8 H2N(CH2)6NH2+8 HOOC(CH2)10COOH→H−[−NH-(CH2)6-NH-CO-(CH2)10-CO−]x−OH+y H2O 8\,\text{H}_2\text{N(CH}_2)_6\text{NH}_2 + 8\,\text{HOOC(CH}_2)_{10}\text{COOH} \rightarrow \text{H}-\left[-\text{NH-(CH}_2)_6\text{-NH-CO-(CH}_2)_{10}\text{-CO}-\right]_x-\text{OH} + y\,\text{H}_2\text{O} 8H2N(CH2)6NH2+8HOOC(CH2)10COOH→H−[−NH-(CH2)6-NH-CO-(CH2)10-CO−]x−OH+yH2ODeduce the values of xxx and yyy in this equation.
Describe how hydrogen bonding occurs between two adjacent sections of the nylon 6,12 polymer, identifying the atoms involved and the geometry of the hydrogen bond.
