{"id":2772,"date":"2018-06-21T13:26:06","date_gmt":"2018-06-21T13:26:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/amines-as-nucleophiles\/"},"modified":"2018-08-09T20:40:20","modified_gmt":"2018-08-09T20:40:20","slug":"9-4-reaction-of-rx-with-nh3-and-amines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/9-4-reaction-of-rx-with-nh3-and-amines\/","title":{"raw":"9.4. Reaction of RX with NH3 and amines","rendered":"9.4. Reaction of RX with NH3 and amines"},"content":{"raw":"
\r\n
\r\n

Preparation of primary amines (RNH2<\/sub>) via the SN2 reaction<\/h1>\r\nAlkyl halides react via SN<\/sub>2 with ammonia (NH3<\/sub>) and amines as the attacking nucleophile. Because these are uncharged nucleophiles, it means that the initial SN<\/sub>2 product has a positive charge on nitrogen, so an additional acid-base step is needed to produce an uncharged product.\r\n\r\n\"\"\r\n\r\n<\/div>\r\n<\/header>
\r\n
\r\n\r\nUnfortunately, the primary amine product is also a powerful nucleophile, and so some of it will attack a second molecule of the alkyl halide.\u00a0 If the primary amine is desired, one way to avoid this is to use a large excess of NH3<\/sub>., so that the reaction rate with NH3<\/sub> greatly exceeds the rate with the amine.\u00a0 In a synthesis question, we would not show the mechanism, and we would simply write the reaction as:\r\n\r\n\"\"\r\n\r\nIn some cases this \"use excess\" approach may be unsuitable, in which case it is common to use a surrogate for the NH3<\/sub> which cannot react further.\u00a0 One traditional approach uses phthalimide, and is called the Gabriel synthesis; a more common approach in modern synthesis is to use sodium azide followed by reduction.\u00a0 Sodium azide (NaN3<\/sub>) is cheap (it is used in air bags!), and the azide ion (N3<\/sub>-<\/sup>) is an excellent nucleophile for SN<\/sub>2.\u00a0 The product is an alkyl azide (RN3<\/sub>), which can be reduced in a separate step to RNH2<\/sub> using excess H2<\/sub>\/Pd or LiAlH4<\/sub>.\u00a0 The mechanism for the second step is complex and not necessary for you to learn at this point.\u00a0 An example synthesis is shown below to illustrate this approach. (You could do this on an exam, but unfortunately alkyl azides with small alkyl groups can be explosive, so I wouldn't recommend running this actual reaction in the lab or on a chemical plant!)\r\n\r\n\"\"\r\n\r\nIn a synthesis question we use (i) and (ii) to show that the reagents are added in two separate steps:\r\n\r\n\"\"\r\n

The reaction of amines with alkyl halides<\/h1>\r\n<\/div>\r\n
\r\n\r\nAs with the NH3<\/sub> reaction described above, when amines are reacted it is common to obtain mixtures because of reactions with more than one molecule of alkyl halide \u00a0 Under some circumstances it may be possible to drive the reaction as was seen in the preparation of primary amines.\u00a0 It is even possible to end up with four alkyl groups on one positively charged nitrogen - a quaternary ammonium salt.\u00a0 For example, consider a reaction between ethylamine (a primary amine) and bromoethane (a primary alkyl halide):\r\n\r\nIn the first stage of the reaction, you get the salt of a secondary amine formed. For example if you started with ethylamine and bromoethane, you would get diethylammonium bromide\r\n\r\n\"\"\r\n\r\nIn the presence of excess ethylamine in the mixture, there is the possibility of a reversible reaction. The ethylamine removes a hydrogen from the diethylammonium ion to give free diethylamine - a secondary amine.\r\n\r\n\"\"\r\n\r\n<\/div>\r\n
\r\n\r\nBut it doesn't stop here! The diethylamine also reacts with bromoethane - in the same two stages as before. This is where the reaction would start if you reacted a secondary amine with a halogenoalkane.\r\n\r\nIn the first stage, you get triethylammonium bromide.\r\n\r\n\"\"\r\n\r\nThere is again the possibility of a reversible reaction between this salt and excess ethylamine in the mixture.\r\n\r\n\"\"\r\n\r\nThe ethylamine removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine - triethylamine.\r\n\r\n<\/div>\r\n
\r\n

Making a quaternary ammonium salt<\/h3>\r\nThe final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).\r\n\r\n\"\"\r\n\r\nThis time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here.\r\n\r\n<\/div>\r\n
<\/div>\r\n<\/section>","rendered":"
\n
\n

Preparation of primary amines (RNH2<\/sub>) via the SN2 reaction<\/h1>\n

Alkyl halides react via SN<\/sub>2 with ammonia (NH3<\/sub>) and amines as the attacking nucleophile. Because these are uncharged nucleophiles, it means that the initial SN<\/sub>2 product has a positive charge on nitrogen, so an additional acid-base step is needed to produce an uncharged product.<\/p>\n

\"\"<\/p>\n<\/div>\n<\/header>\n

\n
\n

Unfortunately, the primary amine product is also a powerful nucleophile, and so some of it will attack a second molecule of the alkyl halide.\u00a0 If the primary amine is desired, one way to avoid this is to use a large excess of NH3<\/sub>., so that the reaction rate with NH3<\/sub> greatly exceeds the rate with the amine.\u00a0 In a synthesis question, we would not show the mechanism, and we would simply write the reaction as:<\/p>\n

\"\"<\/p>\n

In some cases this “use excess” approach may be unsuitable, in which case it is common to use a surrogate for the NH3<\/sub> which cannot react further.\u00a0 One traditional approach uses phthalimide, and is called the Gabriel synthesis; a more common approach in modern synthesis is to use sodium azide followed by reduction.\u00a0 Sodium azide (NaN3<\/sub>) is cheap (it is used in air bags!), and the azide ion (N3<\/sub>–<\/sup>) is an excellent nucleophile for SN<\/sub>2.\u00a0 The product is an alkyl azide (RN3<\/sub>), which can be reduced in a separate step to RNH2<\/sub> using excess H2<\/sub>\/Pd or LiAlH4<\/sub>.\u00a0 The mechanism for the second step is complex and not necessary for you to learn at this point.\u00a0 An example synthesis is shown below to illustrate this approach. (You could do this on an exam, but unfortunately alkyl azides with small alkyl groups can be explosive, so I wouldn’t recommend running this actual reaction in the lab or on a chemical plant!)<\/p>\n

\"\"<\/p>\n

In a synthesis question we use (i) and (ii) to show that the reagents are added in two separate steps:<\/p>\n

\"\"<\/p>\n

The reaction of amines with alkyl halides<\/h1>\n<\/div>\n
\n

As with the NH3<\/sub> reaction described above, when amines are reacted it is common to obtain mixtures because of reactions with more than one molecule of alkyl halide \u00a0 Under some circumstances it may be possible to drive the reaction as was seen in the preparation of primary amines.\u00a0 It is even possible to end up with four alkyl groups on one positively charged nitrogen – a quaternary ammonium salt.\u00a0 For example, consider a reaction between ethylamine (a primary amine) and bromoethane (a primary alkyl halide):<\/p>\n

In the first stage of the reaction, you get the salt of a secondary amine formed. For example if you started with ethylamine and bromoethane, you would get diethylammonium bromide<\/p>\n

\"\"<\/p>\n

In the presence of excess ethylamine in the mixture, there is the possibility of a reversible reaction. The ethylamine removes a hydrogen from the diethylammonium ion to give free diethylamine – a secondary amine.<\/p>\n

\"\"<\/p>\n<\/div>\n

\n

But it doesn’t stop here! The diethylamine also reacts with bromoethane – in the same two stages as before. This is where the reaction would start if you reacted a secondary amine with a halogenoalkane.<\/p>\n

In the first stage, you get triethylammonium bromide.<\/p>\n

\"\"<\/p>\n

There is again the possibility of a reversible reaction between this salt and excess ethylamine in the mixture.<\/p>\n

\"\"<\/p>\n

The ethylamine removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine – triethylamine.<\/p>\n<\/div>\n

\n

Making a quaternary ammonium salt<\/h3>\n

The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide – a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).<\/p>\n

\"\"<\/p>\n

This time there isn’t any hydrogen left on the nitrogen to be removed. The reaction stops here.<\/p>\n<\/div>\n

<\/div>\n<\/section>\n\n\t\t\t
\n\t\t\t

Candela Citations<\/h3>\n\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t\t
CC licensed content, Original<\/div>