Effect of water on the yield of 1-bromobutane?
1. In the procedure, some water was added to the initial reaction mixture. How might the yield of 1-bromobutane be affected by the failure on the part of the student to add the water, and what product(s) would be favored? How might the yield of product be affected by adding, for example, twice as much water as is called for while keeping the quantities of the other reagents the same? 2. In the purification process, the organic layer is washed with 2 M NaOH and then with water. What is the purpose of these washes ?
In the reaction of 3-methyl-2-butanol and HBr, determine the products that would be made.?
Two products will be possible. One more so than the other. 2-bromo-2-methyl butane and 2-bromo-3-methyl butane. The first one will be major. The reasoning behind this is that once you protonate the OH group to form a good leaving group, the H2O group can leave on its own, forming a secondary carbocation. A hydride shift will occur, quickly. Reason being that a shift will allow for the formation of a tertiary carbocation which is a more stable intermediate due to the inductive effect of the adjacent sigma bonds. Br- can then come in to add to the carbocation. The second product will be very minor, if not possible at all. I suppose wants you form the good leaving group, Br- could come in and do an Sn2 reaction. However, it is secondary and sterically crowded so that decreases the chance of Sn2 happening. Experimentation is obviously the best way to answer whether or not we see product 2. Product 1 is certainly expected. I hope this helped.
What happens when 1-Bromo-2-Methylpropane is treated with sodium in the presence of dry ether?
The Wurtz reaction is an organic reaction used to couple two alkyl halides to form an alkane using sodium metal and in presence of dry ether.Let's take the example of 1-Bromo-2-MethylpropaneWhen two molecules of 1-bromo-2-methylpropane are treated with sodium metal in presence of dry ether, 2,5-Dimethylhexane is formed.2 CH3CH(CH3)CH2Br + 2Na → CH3CH(CH3)CH2CH2CH(CH3)CH3 + 2 NaBrThe mechanism begins with a single electron transfer (SET) from sodium metal to the alkyl halide, which dissociates to form an alkyl radical and sodium halide saltRX + Na → R• + NaXThe alkyl radical then accepts an electron from another sodium atom to form carbanion.R• + Na → Na+R-The nucleophilic carbon of the carbanion then displaces the halogen from alkyl halide in an SN2 reaction, forming a new carbon-carbon covalent bond.Na+R- + R-X → R-R + NaX
How do the products differ when ethyl bromide reacts separately with aqueous KOH and alchoholic KOH? Name the product
Hello :) When ethyl bromide reacts with aqueous KOH, it undergoes hydrolysis. This occurs through the SN 2 mechanics since the substrate is a primary carbon. The OH group from KOH substitutes the Br group from ethyl bromide. The by product is KBr. When ethyl bromide reacts with alcoholic KOH, it undergoes beta elimination . The lone pair on O in OH attracts the hydrogen on the beta carbon( the carbon next to the carbon which has the bromide group) and the potasssium cation attracts the bromide group. The product is ethene and the by products are water and KBr.I'm attaching the reactions here as images.Hope it was helpful :)
2-butene is more stable than 1-butene why?
Actually the reason behind it is not related to a sort of symmetry or something !!Hyperconjugation is the effect just like resonance in which the Hydrogen attached to the carbon atom next to double bond increases the stablity of double bond due to bonding in orbitals. 2- Butene having 4 carbon atoms attached with double bonded carbon atoms & each attached carbon have 3 Hydrogen Each . So total 12 no of Hydrogen atoms, while in 1- butene only 1 carbon is attached to double bonded carbon having 2 hydrogens . So the structure with more number of Alpha Hydrogen (Hydrogen attached to carbon atom next to double bond here) is more stablized. so 2-butene with 12 Alpha Hydrogens is more stable than 1-butene with 2 alpha hydrogen.
What is the principle organic product of the reaction of 1-bromopropane and sodium acetate in acetic acid?
To answer this, please do the following steps: 1) separate acetic acid into separate ions (** note where the + and - charges are on the molecule) 2) where would 1-bromopropane react based upon charges in the reaction? This involves knowledge of electron pushing. (**** think: where is electron density coming from and going towards where it lacks?) definitions: 1) electrophile = loves electrons (has none) 2) nucleophile = loves nucleus (has electrons)
How does one convert amphetamine to methamphetamine?
It can easily be done with reductive amination.Alkylating agents can be problematic because secondary amines are more easily alkylated than primary amines. Thus, you end up with a mixture of secondary, tertiary, and quaternary amines. The trick is to do reductive amination. You can easily methylate an amine by reaction with formaldehyde to form an imine. This can be followed by reduction with sodium borohydride, which will yield the desired secondary amine (methamphetamine). It is possible to do it all in one pot if you use sodium cyanoborohydride, which is strong enough to reduce the imine, but too weak to reduce the aldehyde. Thus, as the imine forms the cyanoborohydride reduces it. Imines form via reversible covalent bond formation, which means it's an equilibrium reaction. To drive it to the product, you can do the reaction over 4 angstrom molecule sieves (zeolites), which will absorb any water formed. Alternatively, it could be done in a dean-stark trap where the water is driven off by heat.