10. Cyclohexanone + n-propylmagnesium bromide mechanism 11....

Transcript

00:01 So we have given different parts here first of all we have this ring which is attached to a carbonyl carbon and we have this one two three four so butyl magnesium bromide is there now this magnesium is highly electropositive so that's why it has partial positive charge the carbon which is directly attached to highly electro positive metal that is partially negatively charged this oxygen so this is electron rich here electro negative in nature and that's why it has partial negative charge and the carbon directly attached to the electronegative atom so that will be partial positive charge.

00:41 Okay so now what will happen that this carbon will approach the electron deficient carbon the double bond converts to lone pair of electron and we will have this ring as it is now we have single bond oxygen three non -bonding pair of electrons electron okay of course the negative charge is there okay and we have this alkyre group also with four carbon one two three four okay and this magnesium bromide will be in approach of this then we have ch3 c triple bond ch and nanh2 now let me draw this nitrogen with two hydrogen atoms okay let me open this structure this nitrogen has a pair of electron of course so now here this hydrogen which is directly attached to this triple bondedly carbon atom okay so now this carbon is sp hybridized and that's why this hydrogen will term as acidic hydrogen okay and this is partially nightly charged here okay partially positively charged here and this is partially negatively charged here and now this nitrogen is electron rich okay so now what will happen that this nitrogen being electron rich approaches this electron deficient hydrogen the bond between hydrogen and carbon broken down and we have ch3 c triple bond c and a positive charge negative charge and we have an h3 okay so this hydrogen is with this nitrogen now so that's why this is nh3 here okay then we have third part so we have a ketone like this and we have this cyclohexenring which is directly attached to magnesium bromide okay so again grignard reagent you know this is partially positive charged so this carbon is partial negatively charged this is partial negative charge this is partially positively charged so now what will happen that this carbon will approach the electron deficient carbon and this double bond converts to lone pair of electron so this will be so we have single bond oxygen three unbounded pair of electron negative charge and magnesium bromide will approach this oxygen and also this ring will attach here okay now fourth part so fourth part this is bf3 okay so bf3 can be drawn like this and we have ch3 ohh okay so now this oxygen is electron rich this is partially negatively charged this is partially positively charged and now this boron is electron deficient okay because see it has three electrons in there in its valence shell and all the three electrons are involved in bonding but then two it has six valence and six electron in its octet and to complete the octet it required two more electrons right so that's why bf3 is a electron deficient okay so now what will happen this oxygen being electron rich it will approach the electron deficient boron okay and actually we can say that this is here lavish acid and this will act as levis base okay and they lead to form a product like this let me show you so this bf3 is attached through oxygen to this carbon.

05:35 Okay, so this is the product.

05:39 And of course there will be positive charge because this oxygen has donated its electron two from the bond.

05:47 Fifth part.

05:50 We have nacn, we have this alkali bromide.

05:57 Now what will happen that this bromine is electron negative, this will be partial negative charge here okay and this carbon will have partial positive charge because it is directly a test to a electronegative atom okay so now what will happen here that here this cn this nitrogen so it has a pair of electron being electron rich it will approach the electron deficient carbon so what will happen that this bromine be a good living group it leaves from here okay and that's why we will have a product.

06:38 So now cn will attach here and we have nabr as the bi -product...

Question

10. Cyclohexanone + n-propylmagnesium bromide mechanism 11. CH3C?CH + NaNH2 ? CH3C?CNa + NH3 12. 3-pentanone + cyclopentylmagnesium bromide mechanism 13. BF3 + CH3OH ? F3B-O(H)CH3 14. NaCN + 1-bromopropane ? CH3CH2CH2CN + NaBr

Question

10. Cyclohexanone + n-propylmagnesium bromide mechanism 11. CH3C?CH + NaNH2 ? CH3C?CNa + NH3 12. 3-pentanone + cyclopentylmagnesium bromide mechanism 13. BF3 + CH3OH ? F3B-O(H)CH3 14. NaCN + 1-bromopropane ? CH3CH2CH2CN + NaBr

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