If a tertiary alkanol is oxidized, what is the compound formed?
90884
Tertiary alkanols cannot be readily oxidized since they do not have any hydrogen to be removed from the carbon atom to which the OH group is attached.
225634
None. They are two different compounds.\n\nIt looks like you want to compare a specific property of the alcohols. Please specify!
440879
Alcohols can be prepared by adding water to an alkene in the presence of a strong acid such as concentrated sulfuric acid. This reaction involves adding an H2O molecule across a C=C double bond. Because these reactions follow Markovnikov's rule, the product of the reaction is often a highly substituted 2 or 3 alcohol.
542412
See number 28:\nhttp://elchem.kaist.ac.kr/jhkwak/OkanaganPdb97/molecule/alkane.htm
903981
It depends on the alkyl halide.\n\nA primary alkyl halide would undergo an SN2 reaction with sodium hydroxide because the only time an elimination product is given is with a bulky base like t-butoxide.\n\nA secondary alkyl halide would undergo E1/E2.  If E1, a carbocation intermediate would occur.\n\nA tertiary alkyl halide would undergo E1 with a carbocation intermediate.  No rearrangement would occur because it is already stable.\n\nMy guess is this:  You have a secondary halide that is undergoing E1, which is possible and would result in a rearrangement.\n\nDid you learn about E1?  E1 is basically like SN1, where the rate determining step is the formation of a carbocation intermediate via the leaving of the halide.\n\nI hope this helps.
128824
NaBH4 functions as BH4-, which donates a hydride (a hydrogen with a negative charge) to the aldehyde or ketone that you're working with. The product is a sodium salt/alkoxide. Water or HCl (acid) will protonate the alkoxide to generate the alcohol that you need. \n\nThe gas formed was probably hydrogen, from when the HCl reacted with the Sodium borohydride.
829009
I won't answer this completely, because it looks very much like homework. \n\nI will say that an Aldehyde and a Ketone will yeild different degree alcohols. Reactions that yeild an primary alcohol from an Aldehyde would yeild a secondary alcohol from a ketone. Reactions that yeild an secondary alcohol from an Aldehyde would yeild a tertiary  alcohol from a ketone.
947223
Alcohol, aldehyde, ketone, or carboxylic acid.
876582
Ethanol, or ethyl alcohol.\n\nSplit the water (H2O) into H and OH.\nSplit one of the bonds of the double bond two form two new single bonds.\nAttach the H to one, the OH to the other, to form H3C-CH2OH.
146289
Primary and secondary alcohols react with PCC ('Pyridinium chloride) to form an alhehyde (primary) or a ketone (secondary). The reaction is a softened form of chromium oxidation chemistry. The reaction was discovered in the group of the famous chemist E.J. Corey. The disadvantage is that PCC is toxic and green alternatives to this reaction are definately desired. (C&E news recently reported a non-Tin Barton reduction of an alcohol to an alkyl group, but I digress)
231663
_KCl3 _KCL + _o2? K2Cl4o2 thats the answer i think.:)
345145
O2 is the least soluble in water, followed by Br2.\n\nthe problem is that both LiCl and methanol dissolve completely in water. LiCl since it is an ionic compound, and methanol since it forms hydrogen bonding with water. Therefore any order for LiCl and methanol would be correct I presume.
175104
It is mostly composed of alcohol.  An alcohol/gasoline mix has a lower freezing point than straight gasoline, and mixing in alcohol does not adversely affect the combustion characteristics of gas in most engines.
875082
a. N2O4\nb.PCl5\nc.NF3\nd.S2Cl2
526400
My IUPAC spelling has gone a bit rusty, but here we go:\n\nn-hexanal \nCH3-CH2-CH2-CH2-CH2-CHO\n\n4-methyl-pentanal\nCH(CH3)2-CH2-CH2-CHO\n\n3-methyl-pentanal\nCH3-CH2-CH(CH3)-CH2-CHO\n\n2-methyl-pentanal\nCH3-CH2-CH2-CH(CH3)-CHO\n\n2,3-dimethyl-butanal\nCH(CH3)2-CH(CH3)-CHO\n\n2,2-dimethyl-butanal\nCH3-CH2-C(CH3)2-CHO\n\n2-ethyl-butanal (might be 3-carbonyl-pentane instead?)\n(CH3-CH2)2-CH-CHO
764178
Trick question, multiple answers. . .\n\n1. CH3(CH2)3OH\n2. C2H5CH(OH)CH3\n3. (CH3)2CHCH2OH\n4. (CH3)3COH\n5. (C2H5)2O\n6. CH3(CH2)2OCH3\n7. (CH3)2CHOCH3\n\nOk, #4 is tert-butanol.  All of these are structures for C4H10O\n\ngo to http://www.cem.msu.edu/~reusch/VirtualText/intro3.htm and scroll down to the isomers section.  All the diagrams are listed there.
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)